Difference between revisions of "2001 AIME I Problems/Problem 2"

(Solution)
(Solution)
Line 5: Line 5:
 
Let x be the mean of S. Let a be the number of elements in S.
 
Let x be the mean of S. Let a be the number of elements in S.
 
Then,
 
Then,
<cmath>\FRAC{ax+1}{a+1}=x-13</cmath>
+
<cmath>\frac{ax+1}{a+1}=x-13</cmath> and <cmath>\frac{ax+2001}{a+1}=x+27</cmath>
<cmath>\FRAC{ax+2001}{a+1}=x+27</cmath>
+
<cmath>\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}</cmath>
 +
<cmath>\frac{2000}{a+1}=40</cmath> so <cmath>2000=40(a+1)</cmath>
 +
<cmath>a=49</cmath>
 +
We plug that into our very first formula, and get:
 +
<cmath>\frac{49x+1}{50}=x-13</cmath>
 +
<cmath>49x+1=50x-650</cmath>
 +
<cmath>x=651</cmath>
  
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=2001|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2001|n=I|num-b=1|num-a=3}}

Revision as of 22:03, 21 December 2007

Problem

A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$, and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$. Find the mean of $\mathcal{S}$.

Solution

Let x be the mean of S. Let a be the number of elements in S. Then, \[\frac{ax+1}{a+1}=x-13\] and \[\frac{ax+2001}{a+1}=x+27\] \[\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}\] \[\frac{2000}{a+1}=40\] so \[2000=40(a+1)\] \[a=49\] We plug that into our very first formula, and get: \[\frac{49x+1}{50}=x-13\] \[49x+1=50x-650\] \[x=651\]

See Also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions
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